Date | May 2021 | Marks available | 5 | Reference code | 21M.1.AHL.TZ1.8 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Use l’Hôpital’s rule to find limx→0(arctan 2xtan 3x).
Markscheme
attempt to differentiate numerator and denominator M1
limx→0(arctan 2xtan 3x)
=limx→0(21+4x2)3 sec2 3x A1A1
Note: A1 for numerator and A1 for denominator. Do not condone absence of limits.
attempt to substitute x=0 (M1)
=23 A1
Note: Award a maximum of M1A1A0M1A1 for absence of limits.
[5 marks]
Examiners report
[N/A]